For a particle in uniform circular motion, the acceleration veca at a point P(R, θ) on the circle of radius R is (Here θ is measured from the x-axis)

Question

For a particle in uniform circular motion, the acceleration veca at a point P(R, θ) on the circle of radius R is (Here θ is measured from the x-axis) (A) -(v2/R) cos θ hati+(v2/R) sin θ hatj (B) -(v2/R) sin θ hati+( dotv2/R) cos θ hatj (C) -(v2/R) cos θ hati-(v2/R) sin θ hatj (D) (v2/R) hati+(v2/R) hatj

Tardigrade Admin · Accepted Answer

Clearly veca=ac cos θ(- hati)+ac sin θ(- hatj) =(-v2/R) cos θ hati-(v2/R) sin θ hatj

Q. For a particle in uniform circular motion, the acceleration $a$ at a point $P (R, θ)$ on the circle of radius $R$ is (Here $θ$ is measured from the $x$ -axis)

$- \frac{v ^{2}}{R} cos θ \hat{i} + \frac{v ^{2}}{R} sin θ \hat{j}$

$- \frac{v ^{2}}{R} sin θ \hat{i} + \frac{v ˙ ^{2}}{R} cos θ \hat{j}$

$- \frac{v ^{2}}{R} cos θ \hat{i} - \frac{v ^{2}}{R} sin θ \hat{j}$

$\frac{v ^{2}}{R} \hat{i} + \frac{v ^{2}}{R} \hat{j}$

Clearly
$a = a_{c} cos θ (- \hat{i}) + a_{c} sin θ (- \hat{j})$
$= \frac{- v ^{2}}{R} cos θ \hat{i} - \frac{v ^{2}}{R} sin θ \hat{j}$

Q. For a particle in uniform circular motion, the acceleration a at a point P(R,θ) on the circle of radius R is (Here θ is measured from the x-axis)

−Rv2​cosθi^+Rv2​sinθj^​

−Rv2​sinθi^+Rv˙2​cosθj^​

−Rv2​cosθi^−Rv2​sinθj^​

Rv2​i^+Rv2​j^​